Calculate 1 + 3 + 5 +...+ (2n - 1) for several natural numbers n.

The solutions for 1 + 3 + 5 +...+ (2n - 1), for the first 10 natural numbers are given in the explanation section below.

Step-by-step explanation:

To calculate 1 + 3 + 5 +...+ (2n - 1) for several natural numbers n, we will use several natural numbers to represent n.

For n= 1,

2n - 1 = 2(1) - 1 = 2 - 1 = 1

Hence, 1 = 1

For n = 2,

2n - 1 = 2(2) - 1 = 4 - 1 = 3

Hence, 1 + 3 = 4

For n = 3,

2n - 1 = 2(3) - 1 = 6 - 1 = 5

Hence, 1 + 3 + 5 = 9

For n = 4,

2n - 1 = 2(4) - 1 = 8 - 1 = 7

Hence, 1 + 3 + 5 + 7 = 16

For n = 5,

2n - 1 = 2(5) - 1 = 10 - 1 = 9

Hence, 1 + 3 + 5 + 7 + 9 = 25

For n = 6,

2n - 1 = 2(6) - 1 = 12 - 1 = 11

Hence, 1 + 3 + 5 + 7 + 9 + 11 = 36

For n = 7,

2n - 1 = 2(7) - 1 = 14 - 1 = 13

Hence, 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49

For n = 8,  

2n - 1 = 2(8) - 1 = 16 - 1 = 15

Hence, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64

For n = 9,

2n - 1 = 2(9) - 1 = 18 - 1 = 17

Hence, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81

For n = 10,

2n - 1 = 2(10) - 1 = 20 - 1 = 19

Hence, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

step-by-step explanation:

a rhombus is a parallelogram enjoying his symmetry.

(a) the volume of the cone is 20/3 in3

hence, option a is correct.